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JEE Main 8 Apr 2019 Paper 2 Solved Paper

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Question : 29

Total: 90

The magnetic field of an electromagnetic wave is given by:

=1.6×10−6‌cos(2×107z+6×1015t)(2

The associated electric field will be:

→
E
=4.8×102‌cos(2×107z+6×1015t)(
^
i
−2
^
j
)‌
V
m
→
E
=4.8×102‌cos(2×107z−6×1015t)(−2
^
j
+
^
i
)‌
V
m
→
E
=4.8×102‌cos(2×107z−6×1015t)(2
^
i
+
^
j
)‌
V
m
→
E
=4.8×102‌cos(2×107z+6×1015t)(−
^
i
+2
^
j
)‌
V
m

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